Question: Simplify; express your answer in exponential form. Assume $t\neq 0, r\neq 0$. $\dfrac{{(t^{-1}r^{-2})^{-3}}}{{(t^{2}r^{-5})^{5}}}$
Answer: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(t^{-1}r^{-2})^{-3} = (t^{-1})^{-3}(r^{-2})^{-3}}$ On the left, we have ${t^{-1}}$ to the exponent ${-3}$ . Now ${-1 \times -3 = 3}$ , so ${(t^{-1})^{-3} = t^{3}}$ Apply the ideas above to simplify the equation. $\dfrac{{(t^{-1}r^{-2})^{-3}}}{{(t^{2}r^{-5})^{5}}} = \dfrac{{t^{3}r^{6}}}{{t^{10}r^{-25}}}$ Break up the equation by variable and simplify. $\dfrac{{t^{3}r^{6}}}{{t^{10}r^{-25}}} = \dfrac{{t^{3}}}{{t^{10}}} \cdot \dfrac{{r^{6}}}{{r^{-25}}} = t^{{3} - {10}} \cdot r^{{6} - {(-25)}} = t^{-7}r^{31}$